The COUNT function returns the number of rows in a query.
The syntax for the COUNT function is:
SELECT COUNT(expression)
FROM tables
WHERE predicates;
Note:
The COUNT function will only count those records in which the field in the brackets is
NOT NULL.
For example, if you have the following table called suppliers:
Supplier_ID Supplier_Name State
1 IBM CA
2 Microsoft
3 NVIDIA
The result for this query will return 3.
Select COUNT(Supplier_ID) from suppliers;
While the result for the next query will only return 1, since there is only one row in the
suppliers table where the State field is NOT NULL.
Select COUNT(State) from suppliers;
Simple Example
For example, you might wish to know how many employees have a salary that is above
$25,000 / year.
SELECT COUNT(*) as "Number of employees"
FROM employees
WHERE salary > 25000;
In this example, we've aliased the count(*) field as "Number of employees". As a result,
"Number of employees" will display as the field name when the result set is returned.
Example using DISTINCT
You can use the DISTINCT clause within the COUNT function.
For example, the SQL statement below returns the number of unique departments where
at least one employee makes over $25,000 / year.
SELECT COUNT(DISTINCT department) as "Unique departments"
FROM employees
WHERE salary > 25000;
Again, the count(DISTINCT department) field is aliased as "Unique departments". This
is the field name that will display in the result set.
Example using GROUP BY
In some cases, you will be required to use a GROUP BY clause with the COUNT
function.
For example, you could use the COUNT function to return the name of the department
and the number of employees (in the associated department) that make over $25,000 /
year.
SELECT department, COUNT(*) as "Number of employees"
FROM employees
WHERE salary > 25000
GROUP BY department;
Because you have listed one column in your SELECT statement that is not encapsulated
in the COUNT function, you must use a GROUP BY clause. The department field must,
therefore, be listed in the GROUP BY section.
Performance Tuning
Since the COUNT function will return the same results regardless of what NOT NULL
field(s) you include as the COUNT function parameters (ie: within the brackets), you can
change the syntax of the COUNT function to COUNT(1) to get better performance as the
database engine will not have to fetch back the data fields.
For example, based on the example above, the following syntax would result in better
performance:
SELECT department, COUNT(1) as "Number of employees"
FROM employees
WHERE salary > 25000
GROUP BY department;
Now, the COUNT function does not need to retrieve all fields from the employees table
as it had to when you used the COUNT(*) syntax. It will merely retrieve the numeric
value of 1 for each record that meets your criteria.
Practice Exercise #1:
Based on the employees table populated with the following data, count the number of
employees whose salary is over $55,000 per year.
CREATE TABLE employees
( employee_number number(10) not null,
employee_name varchar2(50) not null,
salary number(6),
CONSTRAINT employees_pk PRIMARY KEY
(employee_number)
);
INSERT INTO employees (employee_number, employee_name, salary)
VALUES (1001, 'John Smith', 62000);
INSERT INTO employees (employee_number, employee_name, salary)
VALUES (1002, 'Jane Anderson', 57500);
INSERT INTO employees (employee_number, employee_name, salary)
VALUES (1003, 'Brad Everest', 71000);
INSERT INTO employees (employee_number, employee_name, salary)
VALUES (1004, 'Jack Horvath', 42000);
Solution:
Although inefficient in terms of performance, the following SQL statement would return
the number of employees whose salary is over $55,000 per year.
SELECT COUNT(*) as "Number of employees"
FROM employees
WHERE salary > 55000;
It would return the following result set:
Number of
employees
3
A more efficient implementation of the same solution would be the following SQL
statement:
SELECT COUNT(1) as "Number of employees"
FROM employees
WHERE salary > 55000;
Now, the COUNT function does not need to retrieve all of the fields from the table (ie:
employee_number, employee_name, and salary), but rather whenever the condition is
met, it will retrieve the numeric value of 1. Thus, increasing the performance of the SQL
statement.
Practice Exercise #2:
Based on the suppliers table populated with the following data, count the number of
distinct cities in the suppliers table:
CREATE TABLE suppliers
( supplier_id number(10) not null,
supplier_name varchar2(50) not null,
city varchar2(50),
CONSTRAINT suppliers_pk PRIMARY KEY
(supplier_id)
);
INSERT INTO suppliers (supplier_id, supplier_name, city)
VALUES (5001, 'Microsoft', 'New York');
INSERT INTO suppliers (supplier_id, supplier_name, city)
VALUES (5002, 'IBM', 'Chicago');
INSERT INTO suppliers (supplier_id, supplier_name, city)
VALUES (5003, 'Red Hat', 'Detroit');
INSERT INTO suppliers (supplier_id, supplier_name, city)
VALUES (5004, 'NVIDIA', 'New York');
INSERT INTO suppliers (supplier_id, supplier_name, city)
VALUES (5005, 'NVIDIA', 'LA');
Solution:
The following SQL statement would return the number of distinct cities in the suppliers
table:
SELECT COUNT(DISTINCT city) as "Distinct Cities"
FROM suppliers;
It would return the following result set:
Distinct Cities
4
Practice Exercise #3:
Based on the customers table populated with the following data, count the number of
distinct cities for each customer_name in the customers table:
CREATE TABLE customers
( customer_id number(10) not null,
customer_name varchar2(50) not null,
city varchar2(50),
CONSTRAINT customers_pk PRIMARY KEY
(customer_id)
);
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7001, 'Microsoft', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7002, 'IBM', 'Chicago');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7003, 'Red Hat', 'Detroit');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7004, 'Red Hat', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7005, 'Red Hat', 'San Francisco');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7006, 'NVIDIA', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7007, 'NVIDIA', 'LA');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7008, 'NVIDIA', 'LA');
Solution:
The following SQL statement would return the number of distinct cities for each
customer_name in the customers table:
SELECT customer_name, COUNT(DISTINCT city) as "Distinct Cities"
FROM customers
GROUP BY customer_name;
It would return the following result set:
CUSTOMER_NAME Distinct Cities
IBM 1
Microsoft 1
NVIDIA 2
Red Hat 3
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